1. Tardigrade
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  4. An ice cube of mass 0.1 kg at 0°C is placed in an isolated container which is at 227°C. The specific heat s of the container varies with temperature T according to the empirical relation s = A + BT, where A = 100 cal kg-1 K-1 and B = 2 × 10-2 cal kg-1. If the final temperature of the container is 27°C, the mass of the container is (Latent heat of fusion of water = 8 × 104 cal kg-1, specific heat of water = 103 cal kg-1 K-1)

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Q. An ice cube of mass $0.1 \,kg$ at $0^{\circ}C$ is placed in an isolated container which is at $227^{\circ}C$. The specific heat s of the container varies with temperature $T$ according to the empirical relation $s = A + BT$, where $A = 100\, cal\, kg^{-1}\, K^{-1}$ and $B = 2 \times 10^{-2}\, cal\, kg^{-1}$. If the final temperature of the container is $27^{\circ}C$, the mass of the container is
(Latent heat of fusion of water $= 8 \times 10^4\, cal\, kg^{-1}$, specific heat of water $= 10^3\,cal\, kg^{-1}\, K^{-1}$)

Thermal Properties of Matter

Solution:

Heat lost by container $=-\int \limits^{300}_{500} m_{C}\left(A+BT\right)dT$
$=-m_{C}\left[AT+\frac{BT^{2}}{2}\right]^{300}_{500}=21600\,m_{C}$
Heat gained by ice $=m_{ice}L+m_{ice}s_{water}\,\Delta T$
$=0.1 \times8 \times 10^{4}+0.1 \times10^{3}\times 27$
$=10700\,cal$
According to principle of calorimetry
Heat lost by container $=$ Heat gained by ice
$21600m_C = 10700$
or $m_C=0.495\,kg$